# Chemistry

posted by .

A titration is performed by adding .600 M KOH to 40.0 mL of .800 M HCl.

Calculate the pH before addition of any KOH.
Calculate the pH after the addition of 5.0 mL of the base.
Calculate the volume of base needed to reach the equivalence point.
What is the pH at the equivalence point (I say 7 because they're both Strong Acid/Strong Base, but not sure).
Calculate the pH after adding 5.00 mL of KOH past the equivalence point.

• Chemistry -

a)HCl --> H^+ + Cl^-
HCl is 100% ionized; therefore, (HCl) = (H^+) = 0.8M
pH = -log(H^+) = ??

b)to start. moles HCl = M x L = 0.800*0.040 = 0.032.
moles KOH added in 5 mL = M x L = 0.600*0.005 = 0.003
................HCl + KOH ==> KCl + H2O
initial......0.032....0........0.....0
change...-0.003...-0.003..+0.003.+0.003
equil.....0.029.....0.....0.003.0.003
So you essentially have 0.029 moles HCl in 45 mL (0.045L) so (HCl) = )H^+) = 0.029/0.045 = 0.644 and pH = -log(H^+) = ??

c) moles HCl to start = M x L = ??
moles KOH needed = same as moles HCl
M KOH = moles KOH/L KOH

d)yes, NaCl is the salt of a strong acid and strong base; therefore, the pH at the equivalence point is 7

e) Do this the same way as the (b) part (the addition of 5 mL KOH) EXCEPT here you are 5 mL past the equivalence point and the principle species in solution is excess KOH. Calculate pOH and pH from that.

## Similar Questions

1. ### chemistry

Calculate the pH at each stage in the titration of 25.0 ml 0.10 M KOH(aq) with 0.25 M HCl(aq). 1. Initiall ( before titration). 2. After addition of 5 ml of HCl. 3. After addition of 9.5 ml of HCl. 4. At the equivalent point. 5. After …
2. ### Chemistry

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.230 M HClO (aq) with 0.230 M KOH (aq). The ionization constant (Ka) for HClO is 4.00 x 10^-8. (b) after addition of 25.0 mL of KOH (d) after addition …
3. ### chem

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HClO(aq) with 0.150 M KOH(aq). The ionization constant for HClO is 4.0*10^-8 a)before addition of any KOH b)after addition of 25.0 mL of KOH c)after …
4. ### Chemistry

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HClO(aq) with 0.200 M KOH(aq). The ionization constant for HClO can be found here. A. Before any addition of KOH B. After addition of 25.0 mL of …
5. ### Chemistry

Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq). pKa1 = 1.3 and pKa2 = 6.7 a) before addition of any KOH b) after addition of 25.0 mL of KOH c) after addition of …
6. ### Chemistry 2

A titration is performed by adding 0.124 M KOH to 40 mL of 0.159 M HNO3. a) Calculate the pH before addition of any KOH. b) Calculate the pH after the addition of 10.26, 25.65 and 50.29 mL of the base.(Show your work in detail for …
7. ### chemistry

The pKb values for the dibasic base B are pKb1 = 2.1 and pKb2 = 7.4. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.75 M B(aq) with 0.75 M HCl(aq). (b) after addition of 25.0 mL of HCl (c) after …
8. ### Chemistry

Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq). (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 50.0 mL of KOH (d) after …
9. ### chemistry

Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.8 M H3PO3(aq) with 1.8 M KOH(aq). before KOH after addition of 25, 50,75,100 ml KOH
10. ### chemistry

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HClO(aq) with 0.150 M KOH(aq). The ionization constant for HClO is 4.0x10^-8. pH before the addition of any KOH?

More Similar Questions